We have:
#int dx/(x(x^2-1)^2)#
Multiply and divide the integrand function by #x#:
#int dx/(x(x^2-1)^2) = int (xdx)/(x^2(x^2-1)^2#
Substitute now: #x^2 = t#; #2xdx = dt#
#int dx/(x(x^2-1)^2) = 1/2 int (dt)/(t(t-1)^2)#
Add and subtract #t# to the numerator:
#int dx/(x(x^2-1)^2) = 1/2 int (1-t+t)/(t(t-1)^2)dt#
using linearity:
#int dx/(x(x^2-1)^2) = 1/2 int t/(t(t-1)^2)dt-1/2 int (t-1)/(t(t-1)^2)dt#
and simplifying:
#int dx/(x(x^2-1)^2) = 1/2 int 1/((t-1)^2)dt-1/2 int 1/(t(t-1))dt#
The first integral can be resolved directly:
#(1) int dx/(x(x^2-1)^2) = -1/2 1/(t-1)-1/2 int 1/(t(t-1))dt#
For the second, we can divide in partial fractions using the same method:
#int 1/(t(t-1))dt = int (1-t+t)/(t(t-1))dt = int (1-t)/(t(t-1))dt + int (tdt)/(t(t-1))#
#int 1/(t(t-1))dt = - int(dt)/t + int (dt)/(t-1)#
#int 1/(t(t-1))dt = -ln abst +lnabs(t-1)+C#
Substituting in the expression #(1)# above:
#int dx/(x(x^2-1)^2) = -1/2 1/(t-1) +1/2lnabs t -1/2 lnabs(t-1)#
and undoing the variable substitution:
#int dx/(x(x^2-1)^2) = -1/2 1/(x^2-1) +1/2lnx^2 -1/2 ln abs(x^2-1)#
and finally, using the properties of logarithms:
#int dx/(x(x^2-1)^2) = -1/2 1/(x^2-1) +ln(absx/sqrt abs(x^2-1))#
